Procedure
First, the lighter before the experiment was measured for comparative purposes after the gas was removed. The jar was then filled with water and then the water was measured while it was dumped out with the graduated cylinder. This was the total volume of the jar. The water temperature was measured with the thermometer for future calculations. Next, the jar was filled again with water all the way to the brim and the cap was screwed on. The jar was then dunked under the tank of water and tipped slightly so that the end of the jar was tilted upwards so the gas did not escape. The lighter was then placed in the water as well, put into the brim of the jar, and the gas was then allowed to escape when the lighter was pressed. The gas then filled the jar about 20% of the way, upon later examination, and the lid was screwed back on while it was still under the water. After the lid was tightly sealed the jar was taken out of the water and allowed to dry. Once the lighter was dry it was measured and the difference was the grams of the substance used in the formula. The jar was then opened and the water was measured with a graduated cylinder. The difference was used for the volume of the gas in the container. After all is measured, the station was cleaned and materials put up to dry.
Data
Original Volume of jar (measured with water, see procedure) - 259 ml
Volume of water in jar after addition of gas- 217.16 ml
Volume of gas (original - after addition of gas, then multiplied by (pressure of gas/standard)) - 40.84 ml
Original Mass of lighter before release of gas- 17 g
Mass of lighter after gas has been released and water drained- 16.9 g
Mass of gas released into jar (original - after) - .1 g
Temperature of water- 23 Celsius or 296 Kelvin
Volume of water in jar after addition of gas- 217.16 ml
Volume of gas (original - after addition of gas, then multiplied by (pressure of gas/standard)) - 40.84 ml
Original Mass of lighter before release of gas- 17 g
Mass of lighter after gas has been released and water drained- 16.9 g
Mass of gas released into jar (original - after) - .1 g
Temperature of water- 23 Celsius or 296 Kelvin
Calculations
PMV=mRT
Pressure- 760 mmHg (standard) - 21 mmHg (water pressure at 23 Celsius)= 739 mmHg (used 739/760 instead so I could use the .0821 constant)
Molar Mass= Is what is being calculated.
Volume= (739/760)(259-217) - 40.84 ml. The pressure of just gas multiplied by the difference in volumes of water. Pressure and Volume are directly related and we're assuming it's at standard which requires extra calculations to correct for that. Divide by 1000 when in equation to convert to L.
m= .1 g. The previous lighter mass (17 g) minus the lighter mass when the gas was released (16.9).
R= Constant. .0821 was used because it was in atm.
T=23 celcius, which was measured. 296 K.
(739/760)(M)(40.84/1000)= (.1)(.0821)(296)
M= 61.2
After solving the equation by plugging in all of the necessary numbers, the M (molar mass) comes out to be 61.2 g/mol.
Pressure- 760 mmHg (standard) - 21 mmHg (water pressure at 23 Celsius)= 739 mmHg (used 739/760 instead so I could use the .0821 constant)
Molar Mass= Is what is being calculated.
Volume= (739/760)(259-217) - 40.84 ml. The pressure of just gas multiplied by the difference in volumes of water. Pressure and Volume are directly related and we're assuming it's at standard which requires extra calculations to correct for that. Divide by 1000 when in equation to convert to L.
m= .1 g. The previous lighter mass (17 g) minus the lighter mass when the gas was released (16.9).
R= Constant. .0821 was used because it was in atm.
T=23 celcius, which was measured. 296 K.
(739/760)(M)(40.84/1000)= (.1)(.0821)(296)
M= 61.2
After solving the equation by plugging in all of the necessary numbers, the M (molar mass) comes out to be 61.2 g/mol.
Conclusion
The calculated molar mass of the gas was 61.2 g/mol. This gave a percent error of 5.299 percent (see analysis question 3 for calculations), meaning that the method used for findng the molar mass was correct. In short, this expirement proved that the molar mass could be derived, as well as a hypothetical element, from the displacement of water in a contained environment. If the gas is also in a contained environment that can be measured for its mass and the water and its container can be weighed as well, there seems to be no limit to calculating the molar mass of the gas in question. While this data is not completely correct, as there are sources of error, it was extremely close to the actual molar mass. Sources of error thatcould have affected the data included water in the lighter, water clingng to the graduated cylinder, gas escaping instead of filling the jar, spilling of any water, and the inaccuracy of the measuring tools (in this case a triple beam).
Anaylsis
1. What is the molar mass of the gas, based of your calculations?
The molar mass based of the calculations was 61.2 g/mol.
2. The gas is an alkane. Based on your molar mass, suggest a possible formula for the gas.
A possible formula for the gas was C4H13. (molar mass of 61.144 g/mol)
3. Most of the gas in the lighter butane. C4H10. Based on that, what is your percent error?
100((61.2/58.12)-1)=5.299. The percent error is calculated to be 5.299%.
4. The effects of possible scenarios.
a) In the calculations for this experiment it would have droped the value down to 4.7 instead of 61.2. This is a a major effect that caused the value to lower substantially. This is because the right side of the equation is much less than it should be creating a smaller amount to divide by.
b) This would result in a slightly lower value because at the end the division that would be done to the right side of the equation would be by a number that is too large. This would cause the number to be somewhat smaller, but not by much as the value with this error was 59.5 compared to the correct value of 61.2.
c) This would cause the molar mass to be smaller, as the volume would be substantially larger. This increase in volume would effect the division of the right side by the left to get M, namely by dividing it by a larger number which would shrink the value. If the difference in volume was 50 ml instead of 40.84 ml the molar mass would be 50 instead of 61.2 g/mol.
d) This would cause a decrease in the grams of gas released, which is the 'm' in the equation. This will cause a decrease in the value of the molar mass because the number would be too small when it came time to divide the right side by the left. If the difference was only .05 g the molar mass calculated would be 30.6 g/mol, which is half.
5. This is actually a mixture of gasses, do they have a higher or lower molar mass?
They have a higher molar mass because the percent error showed that the value recieved was too high. If they were less then the value derived would be too small and would result in a negative value for the percent error. This is assuming that the calculations done were correct and that the molar mass that was calculated wasn't severly impacted by any sources of error that would have created a problem.
6. Explain why this gas is a liquid inside of the lighter.
The gas is a liquid because of the pressure inside of the container. It's impractical to carry around a large amount of gas, so by compressing it it becomes more portable and useful. When the pressure increased the particles are forced closer together, causing a state change from a gas to a liquid. Essentially it's the pressure that causes it to become a liquid, and when the pressure is released it becomes a gas once more.
The molar mass based of the calculations was 61.2 g/mol.
2. The gas is an alkane. Based on your molar mass, suggest a possible formula for the gas.
A possible formula for the gas was C4H13. (molar mass of 61.144 g/mol)
3. Most of the gas in the lighter butane. C4H10. Based on that, what is your percent error?
100((61.2/58.12)-1)=5.299. The percent error is calculated to be 5.299%.
4. The effects of possible scenarios.
a) In the calculations for this experiment it would have droped the value down to 4.7 instead of 61.2. This is a a major effect that caused the value to lower substantially. This is because the right side of the equation is much less than it should be creating a smaller amount to divide by.
b) This would result in a slightly lower value because at the end the division that would be done to the right side of the equation would be by a number that is too large. This would cause the number to be somewhat smaller, but not by much as the value with this error was 59.5 compared to the correct value of 61.2.
c) This would cause the molar mass to be smaller, as the volume would be substantially larger. This increase in volume would effect the division of the right side by the left to get M, namely by dividing it by a larger number which would shrink the value. If the difference in volume was 50 ml instead of 40.84 ml the molar mass would be 50 instead of 61.2 g/mol.
d) This would cause a decrease in the grams of gas released, which is the 'm' in the equation. This will cause a decrease in the value of the molar mass because the number would be too small when it came time to divide the right side by the left. If the difference was only .05 g the molar mass calculated would be 30.6 g/mol, which is half.
5. This is actually a mixture of gasses, do they have a higher or lower molar mass?
They have a higher molar mass because the percent error showed that the value recieved was too high. If they were less then the value derived would be too small and would result in a negative value for the percent error. This is assuming that the calculations done were correct and that the molar mass that was calculated wasn't severly impacted by any sources of error that would have created a problem.
6. Explain why this gas is a liquid inside of the lighter.
The gas is a liquid because of the pressure inside of the container. It's impractical to carry around a large amount of gas, so by compressing it it becomes more portable and useful. When the pressure increased the particles are forced closer together, causing a state change from a gas to a liquid. Essentially it's the pressure that causes it to become a liquid, and when the pressure is released it becomes a gas once more.